Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 121: 17


The solutions are $x=1$ and $x=-\dfrac{3}{4}$

Work Step by Step

$-4x^{2}+x=-3$ Take all terms to the right side of the equation: $0=4x^{2}-x-3$ Rearrange: $4x^{2}-x-3=0$ Factor the left side of the equation: $(4x+3)(x-1)=0$ Using the zero-factor property, set both factors equal to $0$ and solve each individual equation for $x$: $4x+3=0$ $4x=-3$ $x=-\dfrac{3}{4}$ $x-1=0$ $x=1$ The solutions are $x=1$ and $x=-\dfrac{3}{4}$
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